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Question
A man on the top of vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower? (Give your answer correct to nearest seconds).
Solution
Here, ∠ACB = 30° and ∠ADB = 45°.
Let C denote the initial position of the car and D be its position after 12 minutes.
Let the speed of the car be x meter/minute, then
CD = 12x meters .....( ∵ Distance = speed x Time)
Let the car take t minutes to reach the tower from D.
Then, DB = tx meters
Now in the right-angled triangles ACB,
tan 30° = `"AB"/"BD"`
⇒ `1/sqrt3 = "AB"/("BC" + "CD")`
⇒ `1/sqrt3 = "AB"/(12x + tx)`
⇒ `"AB" = (12x + tx)/sqrt3` ....(1)
Also, in the right-angled triangle ADB,
tan 45° = `"AB"/"DB"`
⇒ 1 = `"AB"/"DB"`
⇒ AB = DB = tx ......(2)
From (1) and (2), we have
t = `12/(sqrt3 - 1) = 12(sqrt3 + 1)/2`
t = `6(sqrt3 + 1)`
t = 15.39
∴ Time = 16.39 minutes
Time = 16 minutes 23 seconds.
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